Question: As a particle moves along the number line, its position at time $t$ is $s(t)$, its velocity is $v(t)$, and its acceleration is $a(t)=-\cos t$. If $v(\pi) = 2$ and $s(\pi/2) = 3\pi$, what is $s(0)$ ? $s(0)=~$
The integral of $~a(t)~$ is $~~v(t)=\int (-\cos t)\,dt=-\sin t+C\,$. We are given that $~v(\pi) =2\,$, so $~2=-\sin\pi+C=0+C\,$. That gives us $~C=2\,$. Therefore, $~v(t) = -\sin t+2\,$. The integral of $~v(t)~$ is $~~s(t)=\int\big(-\sin t +2\big)\,dt=\cos t +2t+K\,$. We are given that $~s\Big(\dfrac\pi2\Big) =3\pi\,$, so $~3\pi=\cos\dfrac\pi2+2\cdot\dfrac\pi2+K=\pi+K\,$. That gives us $ ~K =2\pi\,$. Therefore $~s(t)=\cos t+2t+2\pi\,$. Finally, we can evaluate $~s(0)\,$. $ s(0)=\cos 0+2\cdot0+2\pi=1+2\pi\,$